Question

(x+y)^4+(x^2-y^2)^2+(x-y)^4 因式分解

用“换元法”解出

Answer 1

(x+y)^4+(x^2-y^2)^2+(x-y)^4
=(x+y)^4+(x+y)^2(x-y)^2+(x-y)^4
=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=[(2x^2+2y^2)+(x^2-y^2)][(2x^2+2y^2)-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)

用换元法的话，就把x+y=a.x-y=b，代入，最后回代即可